# Limits: paths of approach

1. Feb 25, 2014

### reddawg

1. The problem statement, all variables and given/known data
By considering different paths, show that the given function has no limit
as (x,y) $\rightarrow$ (0,0).

f(x,y) = x4/(x4 + y4)

2. Relevant equations

3. The attempt at a solution
My instructor taught me this process a while back and am unsure if it fits for this problem:

let y=mx2

limit as (x, mx2) $\rightarrow$ (0,0) of
x4/(x4 + mx4)

I tried this method seeing as letting x=0 yields limit = 0 and letting y = 0 yields limit = 1 ???

2. Feb 25, 2014

### Staff: Mentor

Your substitution is incorrect. If y = mx2, then y4 ≠ mx4.

3. Feb 25, 2014

### reddawg

Oh right, it would become m2x4

Therefore it's x4/(x4 + m2x4)

Factoring out x4 yields 1/(1+m2)

In similar problems this method proved that the limit depended on the value of m, therefore it did not exist. Is this the case here? I would think so.

4. Feb 25, 2014

### HallsofIvy

Staff Emeritus
Yes, different values of that limit for different values of "m" mean that the limit depends upon the path. Recall that if a function has a limit then the value of f must be close to that limit. But this shows the value of f close to (0,0) on the line y=2x is different from the value on the line y= 3x.

Last edited: Feb 25, 2014
5. Feb 25, 2014

### Staff: Mentor

No, that's wrong as well.
If y = mx2 (as you have in the OP), then y4 = (mx2)4, right?

6. Feb 25, 2014

### reddawg

Actually, that was a typo on my part while identifying the problem statement. It is supposed to be
y2 not y4.

Thank you for taking the time to catch that mistake though.